How many moles of precipitate will be formed when 140.0 mL of 0.250 M LiCl is reacted with excess Pb(NO₃)₂ in the following chemical reaction? 2 LiCl(aq) + Pb(NO₃)₂(aq) → PbCl₂(s) + 2 LiNO₃(aq)

Answer 1

2LiCl(aq) + Pb(NO₃)₂(aq) ==> PbCl₂(s) + 2LiNO₃(aq) ... balanced equation

Since Pb(NO₃)₂ is in excess, we need only concern ourselves with moles of LiCl present.

molar mass LiCl = 42.39 g/mol

moles LiCl present = 140.0 ml x 1 L/1000 ml x 0.250 mol/L = 0.0350 moles LiCl

moles PbCl₂ theoretically formed = 0.0350 mol LiCl x 1 mol PbCl₂ / 2 mol LiCl = 0.0175 mol PbCl₂ formed